Concrete OperatorsResearch Article • DOI: 10.2478/conop-2012-0002 • CO • 2013 • 11-18

On the Normality of the Unbounded Product of TwoNormal Operators∗

AbstractLet A and B be two -non necessarily bounded- normal oper-ators. We give new conditions making their product normal.We also generalize a result by Deutsch et al on normal prod-ucts of matrices.

KeywordsUnbounded Operators• Normal Operators • Fuglede-PutnamTheorem

MSC: 47A05, 47B25© Versita sp. z o.o.

Mohammed Hichem Mortad† ‡

Department of Mathematics, University of Oran, B.P. 1524,El Menouar, Oran 31000, Algeria

Received 5 July 2012Accepted 18 September 2012

1. IntroductionFirst, we assume the reader is very familiar with notions, definitions and results on unbounded operators. All unboundedoperators are assumed to be densely defined. Some general references are [1, 4, 8, 18, 19]. We just recall that anunbounded operator A is said to be normal if it is closed and AA∗ = A∗A. We also note that between operators, thesymbol "⊂" stands for extensions, i.e. A ⊂ B means that Ax = Bx for all x ∈ D(A) and that D(A) ⊂ D(B).The question of when the product of two normal operators is normal is fundamental. For papers dealing with boundednormal products, see e.g. [7, 9, 17, 20, 21]. See also the recent paper [3] and the references therein for the boundedoperators case. For the unbounded case, see [13, 15]. For closely related topics see [10, 11]. For those interested insums of normal operators, see [12] and [16].The following example illustrates that the passage from the bounded case to the unbounded one needs care.Example 1. Let A be an unbounded normal operator having a trivial kernel, for example take Af (x) = (1 + x2)f (x) onD(A) = {f ∈ L2(R) : (1 + x2)f ∈ L2(R)}. Note that A is one-to-one but with properly dense range.Now set B = A−1. Observe that both A and B are normal on their respective domains (they are even self-adjoint andpositive!). However BA, defined on D(BA) = D(A), is not closed as BA ⊂ I . Thus it cannot be normal and yet B doescommute with A.For the reader’s convenience, let us summarize, in a chronological order, all what has been obtained, to the author bestknowledge, as regards to the unbounded normal product of two operators:Theorem 1 ([13]).

1. Assume that B is a unitary operator. Let A be an unbounded normal operator. If B commutes with A (i.e. BA ⊂ AB),then BA is normal.2. Assume that A is a unitary operator. Let B be an unbounded normal operator. If A commutes with B (i.e. AB ⊂ BA),then BA is normal.

∗ Supported, in part, by Laboratoire d’Analyse Mathématique et Applications.† E-mail: mhmortad@gmail.com, mortad@univ-oran.dz.‡ Dr Mohammed Hichem Mortad, BP 7085 Seddikia, Oran 31013, Algeria.

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Dropping the unitarity hypothesis the following three results (also in [13]) were obtained:Theorem 2.Let B be a bounded normal operator. Let A be an unbounded normal operator. Assume that B commutes with A. If forsome r > 0, ||rBB∗ − I|| < 1, then BA is normal if it is closed.Theorem 3.Let B a bounded normal operator and let A be an unbounded normal operator which commutes with B. Assume that forsome r > 0, ||rBB∗ − I|| < 1. Then AB is normal.Remark. Observe that the last two results generalize Theorem 1.Proposition 1.Let A be an unbounded normal operator and let B be a bounded normal operator commuting with A. If BB∗ is stronglypositive (in the sense given in [5]), then BA is normal.Very recently, in the context of generalizing Kaplansky’s theorem (see [7]) one finds the following result. Of course, anassumption of unitarity on one of the operators is a strong one.Theorem 4 ([15]).If A is unitary and B is an unbounded normal operator, then

BA is normal ⇐⇒ AB is normal.In the present paper, we obtain new results by assuming that AB = BA in lieu of BA ⊂ AB, under the conditions A andB both normal where only one of them is bounded. Then we show that an anti-commuting relation also gives a similarresult. Then we show that in Theorem 2, the closedness of BA is not needed. Then we generalize a result by Deutschet al which appeared in [2] to unbounded operators. Finally, we establish the normality of the product AB where bothoperators are unbounded.To prove most of the results, we will make use of the following well-known results.Lemma 1.[[8],[18]] If B is (unbounded) symmetric and A is self-adjoint, then

A ⊂ B =⇒ A = B.

Lemma 2.[[8],[18]] If A is closed, then A∗A and AA∗ are both self-adjoint.Corollary 1.If A is a closed operator such that AA∗ ⊂ A∗A, then A is normal.Lemma 3 ([6] or [19]).If A and B are densely defined and A is invertible with inverse A−1 in B(H), then (BA)∗ = A∗B∗.It is known that if B is bounded and A1 and A2 are unbounded and normal, then

BA1 ⊂ A2B =⇒ BA∗1 ⊂ A∗2B.

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This is the well-known Fuglede-Putnam theorem. We can also derive the following version (also known but we includea proof for the reader’s convenience) :Theorem 5.If B is bounded and A1 and A2 are unbounded and normal, then

BA1 = A2B =⇒ BA∗1 = A∗2B.

Proof. By the Fuglede-Putnam theorem we haveBA1 = A2B =⇒ BA1 ⊂ A2B =⇒ BA∗1 ⊂ A∗2B.

Hence BA∗1 = A∗2B forD(A∗2B) = D(A2B) = D(BA1) = D(A1) = D(A∗1) = D(BA∗1).

A recently obtained generalization of the Fuglede-Putnam theorem is also valuable. It readsTheorem 6 (Fuglede-Putnam-Mortad).Let A be a closed operator with domain D(A). Let M and N be two unbounded normal operators with domains D(N)and D(M) respectively. If D(N) ⊂ D(AN), then

AN ⊂ MA =⇒ AN∗ ⊂ M∗A.

2. New ResultsNext theorem is our first resultTheorem 7.Let A and B be two normal operators. Assume that B is bounded. If BA = AB, then BA (and so AB) is normal.Proof. Since BA = AB, by Theorem 5 we have BA∗ = A∗B. Then we have

(BA)∗BA = A∗B∗BA = A∗B∗AB ⊂︸︷︷︸classic FugledeA∗AB∗B

andBA(BA)∗ = BAA∗B∗ = ABA∗B∗ = AA∗BB∗ = A∗AB∗B.

Whence (BA)∗BA ⊂ BA(BA)∗.But BA is closed for it equals AB which is closed since A is closed and B is bounded. Therefore, BA(BA)∗ and (BA)∗BAare both self-adjoint (by Lemma 2) and hence BA is normal (by Corollary 1), completing the proof.Remark. The assumption AB ⊂ BA cannot merely be dropped. By Example 1,

D(AB) = L2(R) 6⊂ D(BA) = D(A) = {f ∈ L2(R) : (1 + x2)f ∈ L2(R)}.

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We also obtain an "anti-commuting version" of Theorem 7.Theorem 8.Let A and B be two normal operators. Assume that B is bounded. If BA = −AB, then BA (and so AB) is normal.Proof. The same idea of proof as that of the previous result applies. We have BA∗ = −A∗B thanks to Theorem 5because −A is also normal. Then

(BA)∗BA = A∗B∗BA = −A∗B∗AB ⊂︸︷︷︸FugledeA∗AB∗B

andBA(BA)∗ = BAA∗B∗ = −ABA∗B∗ = AA∗BB∗ = A∗AB∗B.

The rest is obvious.Now, we improve Theorem 2 by removing the assumption that BA be closed.Theorem 9.Let B be a bounded normal operator. Let A be an unbounded normal operator. Assume that B commutes with A. If forsome r > 0, ||rBB∗ − I|| < 1, then BA is normal.Proof. The proof is the same as the one in [13]. What we are concerned with here is to show that the closedness ofBA is tacitly assumed.So let us show that BA is closed. Let xn → x and BAxn → y. The condition ||rBB∗ − I|| < 1, plus the normality of B,guarantees that BB∗ = B∗B is invertible. Hence, by the continuity of B∗, B∗BAxn → B∗y. Therefore,

Axn −→ (B∗B)−1B∗y.

But A is closed, hence x ∈ D(A) and Ax = (B∗B)−1B∗y. This implies thatB∗BAx = B∗y and hence BB∗BAx = BB∗y

which, thanks to the invertibility of BB∗, clearly yields BAx = y, proving the closedness of BA.Next, we give an unbounded oeprator version of a result by Deutsch et al in [2] (cf. [20] and [21]) on normal products ofmatrices. We haveTheorem 10.Let A be a bounded and invertible operator. Let B be unbounded and closed. Assume further that D(B) ⊂ D(BAB).Then BA and AB are normal iff BAA∗ = A∗AB and B∗BA ⊂ ABB∗.Proof. First, we note that we should not worry about the closedness of both BA and AB for the boundedness andthe invertibility of A (and the closedness of B!) implies that BA and AB are closed respectively.

1. Assume that BAA∗ = A∗AB and B∗BA ⊂ ABB∗ and let us show that BA and AB are normal. Since A is invertible,Lemma 3 implies that (BA)∗ = A∗B∗, and alsoB∗BA ⊂ ABB∗ =⇒ BB∗A∗ ⊂ A∗B∗B,

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where we also used Lemma 2 for B. Hence(BA)∗BA = A∗B∗BA ⊃ BB∗A∗A.

So by using again the invertibility of A (and hence that of A∗A) and Lemma 2 we obtain(BB∗A∗A)∗ = A∗ABB∗ ⊂ ((BA)∗BA)∗ = (BA)∗BA.

On the other hand, we see thatBA(BA)∗ = BAA∗B∗ = A∗ABB∗

which implies thatBA(BA)∗ ⊂ (BA)∗BA.

Corollary 1 then makes the "inclusion" an exact equality, i.e. establishing the normality of BA.Let us turn now to the product AB. This is more straightforward. We have(AB)∗AB = B∗A∗AB = B∗BAA∗

andAB(AB)∗ = ABB∗A∗ ⊃ B∗BAA∗.

Arguing similarly as before gives the normality of AB. This finishes the first part of the proof.2. Assume that BA and AB are both normal. Then

A(BA) = (AB)A =⇒ A(BA)∗ = (AB)∗A =⇒ AA∗B∗ = B∗A∗A

by Theorem 5 and the invertibility of A.We also haveB(AB) = (BA)B =⇒ B(AB) ⊂ (BA)B =⇒ B(AB)∗ ⊂ (BA)∗B

by Theorem 6 (since D(B) ⊂ D(BAB)) and the boundedness of A. HenceBB∗A∗ ⊂ A∗B∗B or B∗BA ⊂ ABB∗

and the proof is then complete.Consider next the following example:Example 2. Let A and B be the two operators defined by

Af (x) = eixf (x) and Bf (x) = ex2−ixf (x)on their respective domains

D(A) = L2(R) and D(B) = {f ∈ L2(R) : ex2 f (x) ∈ L2(R)}.

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Then A is unitary (so BAA∗ = A∗AB is verified) and B is normal. Moreover, we can easily check that:D(B∗BA) = {f ∈ L2(R) : e2x2 f (x) ∈ L2(R)}

andD(ABB∗) = D(BB∗) = {f ∈ L2(R) : e2x2 f (x) ∈ L2(R)}

too. SinceB∗BAf (x) = ABB∗f (x), ∀f ∈ D(B∗BA) = D(ABB∗),

we have B∗BA = ABB∗. We also see that both AB and BA are normal on their equal domainsD(AB) = D(BA) = {f ∈ L2(R) : ex2 f (x) ∈ L2(R)}

since they are the multiplication operator by the function ex2 . Nonetheless we haveD(BAB) = {f ∈ L2(R) : e2x2 f (x) ∈ L2(R)}

and so D(B) 6⊂ D(BAB) as, for instance, e− 32 x2 ∈ D(B) but e− 32 x2 6∈ D(BAB).This example suggests that replacing "bounded and invertible" by "unitary" might allow us to drop the condition D(B) ⊂D(BAB) there. This is in fact the case and we haveTheorem 11.Let A be a unitary operator. Let B be unbounded and closed. Then BA and AB are normal iff B∗BA ⊂ ABB∗.Proof. The proof of sufficiency is as before. Note that with A assumed unitary, the first condition of Theorem 10 isautomatically satisfied.Let us suppose that BA and AB are both normal and let us check that B∗BA ⊂ ABB∗. In fact, since AB is normal, wehave (AB)∗AB = B∗A∗AB = B∗B = AB(AB)∗ = ABB∗A∗.

Hence BB∗A∗ = A∗B∗B. Accordingly by taking adjoints,ABB∗ = B∗BA,

establishing the result.We now turn to the case of two unbounded normal operators. We haveTheorem 12.Let A be an unbounded invertible normal operator. Let B be an unbounded normal operator. If BA = AB, A∗B ⊂ BA∗and B∗A ⊂ AB∗, then BA is normal.Proof. Since A is invertible, by Lemma 3, (BA)∗ = A∗B∗. Then

(BA)∗BA = A∗B∗BA = A∗B∗AB ⊂ A∗AB∗B

andBA(BA)∗ = BAA∗B∗ = BA∗AB∗ ⊃ A∗BAB∗ = A∗ABB∗.

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Therefore, (BA)∗BA ⊂ BA(BA)∗.Since BA = AB, A is invertible and closed, and B is closed, BA is closed and Lemma 1 does the remaining job, i.e.gives us: (BA)∗BA = BA(BA)∗,completing the proof.The same method of proof yieldsTheorem 13.Let A be an unbounded invertible normal operator. Let B be an unbounded normal operator. If BA ⊂ AB, A∗B ⊂ BA∗and B∗A ⊂ AB∗, then BA is normal whenever it is closed.Finally, adopting the same idea of the proof of Theorem 12 and using Theorem 6, we can impose some conditions ondomains to derive a domains-dependent version of Theorem 1.Corollary 2.Let A and B be two unbounded invertible normal operators with domains D(A) and D(B) respectively. If BA = AB andD(A), D(B) ⊂ D(BA), then BA (and AB) is normal.Proof. Note first that the closedness of BA is clear. Now we have

BA ⊂ AB =⇒ BA∗ ⊂ A∗B =⇒ B∗A ⊂ AB∗

by D(A) ⊂ D(BA), Theorem 6 and the invertibility of A∗. Similarly, we haveAB ⊂ BA =⇒ AB∗ ⊂ B∗A =⇒ A∗B ⊂ BA∗

by D(B) ⊂ D(AB), Theorem 6 and the invertibility of B∗. So we came back to the setting of Theorem 12.3. ConclusionNew results for the normality of the unbounded product of two normal operator, have been obtained. A result by Deutschet al for normal matrix products has been generalized to general and unbounded products.References

[1] Conway J.B., A Course in functional analysis, Springer, 1990 (2nd edition)[2] Deutsch E., Gibson P.M., Schneider H., The Fuglede-Putnam theorem and normal products of matrices. Collectionof articles dedicated to Olga Taussky Todd, Linear Algebra and Appl., 1976, 13/1-2, 53-58[3] Gheondea A., When are the products of normal operators normal? Bull. Math. Soc. Sci. Math. Roumanie (N.S.),2009, 52(100)/2, 129-150[4] Goldberg S., Unbounded linear operators, McGraw–Hill, 1966[5] Gustafson K., Positive (noncommuting) operator products and semi-groups, Math. Z., 1968, 105, 160-172[6] Gustafson K., On projections of selfadjoint operators, Bull. Amer. Math Soc., 1969, 75, 739-741[7] Kaplansky I., Products of normal operators, Duke Math. J., 1953, 20/2, 257-26017

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[8] Kato T., Perturbation theory for linear operators, 2nd Edition, Springer, 1980[9] Kittaneh F., On the normality of operator products, Linear and Multilinear Algebra, 1991, 30/1-2, 1-4[10] Mortad M.H., An application of the Putnam-Fuglede theorem to normal products of self-adjoint operators, Proc.Amer. Math. Soc., 2003, 131/10, 3135-3141[11] Mortad M.H., On some product of two unbounded self-adjoint operators, Integral Equations Operator Theory, 2009,64/3, 399-408[12] Mortad M.H., On the normality of the sum of two normal operators, Complex Anal. Oper. Theory, 2012, 6/1, 105-112.DOI: 10.1007/s11785-010-0072-7[13] Mortad M.H., On the closedness, the self-adjointness and the normality of the product of two unbounded operators,Demonstratio Math., 2012, 45/1, 161-167[14] Mortad M.H., An all-unbounded-operator version of the Fuglede-Putnam theorem, Complex Anal. Oper. Theory, (inpress). DOI: 10.1007/s11785-011-0133-6[15] Mortad M.H., Products of Unbounded Normal Operators. arXiv:1202.6143v1[16] Mortad M.H., The Sum of Two Unbounded Linear Operators: Closedness, Self-adjointness and Normality, (submit-ted). arXiv:1203.2545v1[17] Patel A., Ramanujan P.B., On sum and product of normal operators, Indian J. Pure Appl. Math., 1981, 12/10,1213-1218[18] Rudin W., Functional analysis, McGraw-Hill, 1991 (2nd edition)[19] Weidmann J., Linear operators in Hilbert spaces, Springer, 1980[20] Wiegmann N.A., Normal products of matrices, Duke Math J., 1948, 15, 633-638[21] Wiegmann N.A., A note on infinite normal matrices, Duke Math. J., 1949, 16, 535-538

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